\magnification=\magstep 1\def\frac#1#2{{{#1}\over{#2}}}\parindent=10 pt\parskip=8 pt
\input psfig
\centerline{Exam 2 Math 308, S98}

Most of this exam is closed-book and closed-computer. Finish the closed book part before logging on.You will  have to use Maple for problem 6. Where a question with a yes or no answer is set, you get positive points for a correct answer, 1 point for leaving it blank, and negative points for a wrong answer. 

\item{1.} Two boxed pizzas, one large pizza at 200 degrees, and one medium at 160 degrees, were put in a perfect insulating wrap and delivered. When the pizza delivery guy got to the door ten minutes later, the two had temperatures of 195 degrees and 170 degrees respectively. But the folks who ordered the pizza were not ready and it took five minutes to come up with the money. How hot was the large pizza when they finally got it in their hands? 
\item{1S} First pizza is twice as large as the second, in terms of heat capacity, because the little one gained twice the temperature the big one lost. So the total PizzaThermalUnits in the system is $200+160/2=560/3$. (1 unit heats the big pizza 1 degree). Now the temp of the little pizza will be 2*(280-y) so it can have heat 280-y when temp of the big pizza is y. So the DE is $ y'=K*(560-2y-y)=K*(Little pizza temp- big pizza temp).$
Substitute $z=y-560/3$ or just grind it out. By grindout, $y'+3Ky=560K\Rightarrow y\exp(3Kt)=(560/3)\exp(3Kt)+C$ and then $y=560/3+C\exp(-3Kt)$. Setting $t=0$ gives $C=40/3$ so that $y(0)=200$. Then setting $t=10$ you need $560/3+(40/3)\exp(-30K)=195$ so $\exp(-30K)=25/40$. Now setting $t=15$ gives the answer, since $C$ and $K$ are now known. At $t=15$,  $y(15)=560/3+(40/3)\cdot(25/40)^{3/2}$. (Approximately 193).  

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\item{2.} The graphic below shows two functions $f_1$ and $f_2$, which may or may not be solutions to the differential equation $y''+(1/x)y'+y=0$. $$\psfig{figure=test2fig.eps,height=1.9 in}$$
\itemitem{(a)} Explain why the Wronskian of $f_1$ and $f_2$ must be negative at $x=A$ and positive at $x=B$. Because $f_1(A)$ is positive, $f_2'(A)$ is negative, and the other factor in the Wronskian is zero  since $f_1'(A)=0$, while at $B$, $f_1$ is positive, $f_2'$ is positive, and again the other factor is zero for the same reason. 
\itemitem{(b)} Given that the Wronskian of $f_1$ and $f_2$ is negative at $x=A$ and positive at $x=B$, does it follow that they cannot both be solutions to that differential equation? YES The Wronskian to a homogeneous differential equation does not change sign over an interval where the functions $p$ and $q$ are continuous. Here, $1/x$ is continuous from $A$ to $B$. 
\itemitem{(c)} Could both these functions be solutions to $y''+(1/x)y'+y=x$? NO At $x=10$, $y''+(1/10)y'+y=10$ is not even close to being satisfied by either $f$. 
\itemitem{(d)} Could they both be solutions to $y''+(1/x)y'+y=1$? YES. The constant function $y=1$ is a particular solution to the differential equation $y''+(1/x)y'+y=1$. These oscillate about 1, so if $y_p=1$ then $f_1-1$ and $f_2-1$ would oscillating about zero as homogeneous solutions can do. 
\itemitem{(e)} Could they both be solutions to $y''+(1/x)y'+y=2$? NO. Same objection. Subtracting 2 from both $f_1$ and $f_2$ (and the constant 2 is clearly a solution to the differential equation $y''+(1/x)y'+y=2$) would give a homogeneous DE where the Wronskian condition was violated by $f_1-2$ and $f_2-2$. The Wronskian would change sign from $A$ to $B$. 

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\item{3.} Solve $y''+2y'+2y=t$ with $y(0)=y'(0)=0$. $y=(t-1)/2+(1/2)e^{-t}\cos t$.
\item{4.} Exponentials and trig:
\itemitem{(a)}State the algebraic rule connecting complex exponentials to trigonometry. It's $e^{iz}=\cos z+i\sin z$. 
\itemitem{(b)} Without using this rule, but just using the rules of calculus for $e^{ct}$, prove that $f(t):=(1/2)(e^{it}+e^{-it})$ satisfies $f''=-f$. Well, $f'=(1/2)(ie^{it}+(-i)e^{-it})$ and $f''=(1/2)(i^2e^{it}+(-i)^2e^{-it})$. But since both $i^2=-1$ and $(-i)^2=-1$ this gives $-f$. 
\itemitem{(c)} Using also the rules of algebra for $e^0$, prove that $f(0)=1$ and $f'(0)=0$. First, $e^0=1$ so $f(0)=(1/2)(1+1)=1$. Next, $f'(0)=(1/2)(ie^0+(-i)e^{-0})=0$. 
\itemitem{(d)} Both $f(t)$ and $\cos t$  satisfy the same second order linear differential equation with continuous coefficients. Both are equal at zero, and they have the same derivative at zero as well. On the basis of the existence and uniqueness theorem, does this prove that $f(t)=\cos t$ for all $t$? YES. If two functions are equal with equal derivative at some point, and if they both satisfy the same second order linear differential equation, then the functions remain equal for as long as the coefficients in the differential equation are continuous. Here, that is the whole interval from $-\infty$ to $\infty$. 
\itemitem{(e)} The amplitude of the sum wave $3\cos t+4\sin t$ is larger than 4, but less than 8. What is the amplitude? The amplitude is $\sqrt{3^2+4^2}=5$. \vfill\eject

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\item{5.} A 2 kg mass is suspended from a spring with spring constant 100 (all in MKS units). 
\itemitem{(a)} How much will the spring sag under the weight of this mass? Sags $2g/100$ meters, or $2g$ centimeters. Here on earth, $g=9.8$ so that would be a $19.6$ cm sag. But you didn't have to know $g$ to give the other version of the answer. 
\itemitem{(b)} What is the diffferential equation governing the motion of this spring-mass system, assuming no friction or damping? Measure distance from the point at which the mass rests when hanging from the spring. $my''+ky=0$ here becomes $2y''+100y=0$. 
\itemitem{(c)} If there is a little bit of damping, which aspect of the response will be most affected: the ratio of the amplitude of the motion from one cycle to the next, or the time required to complete one cycle? The amplitude ratio is affected much more strongly. A little bit of damping has almost no effect on the natural frequency of a system. And a good thing it is too, because this means that Grandfather Clocks are not particularly thrown off by having the pendulum swing through a bit of air resistance.  

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\item{6.} Give Maple code for a superimposed plot of the solutions to $y''-c\cos(y')+\sin y=0$ with two initial states: $y(0)=0$ and $y'(0)=1$, and $y(0)=1$ and $y'(0)=0$, over the interval $0\le x\le 10$. Here, $c$ is the quantity (YOUR STUDENT ID NUMBER)*$10.0^{-10}$. Write down your coefficient $c$, and sketch the plot on your paper.. 
with(DEtools);

DEplot(diff(y(t),t\$2)=c*cos(diff(y(t),t))-sin(y),y(t),t=0..5,

[[y(0)=0,D(y)(0)=1],[y(0)=1,D(y)(0)=0]],stepsize=0.02,linecolor=black) 

should do. You don't really have to have the bit about stepsize and about linecolor but they don't hurt. 
\bye